Symmetry and the Fourth Dimension (Part 13)

Now let’s start thinking about 4d Platonic solids. We’ve seen the 4-cube… what else is there? Well, in 3d we can take a cube and build an octahedron as shown here. The same trick works in any dimension. In n dimensions, we get something called the n-dimensional cross-polytope, which has one corner at the center of each (n-1)-dimensional face of the n-cube.

Puzzle 1. What’s a 2d cross-polytope?

It’s worth noting the relationship between cubes and cross-polytopes is symmetrical. In other words, we can also build an n-cube by putting one corner at the center of each (n-1)-dimensional face of the n-dimensional cross-polytope! For example:

It gets tiring to say ‘(n-1)-dimensional face’, so people use the word facet for an (n-1)-dimensional face of an n-dimensional thing.

Now let’s think a bit more carefully about what happens in 4 dimensions. Since the 4-cube has 8 facets (each being a cube), the 4d cross-polytope must have 8 corners. And since the 4-cube has 16 corners, the 4d cross-polytope must have 16 facets. This is why it’s also called the 16-cell.

In four dimensional geometry, a 16-cell is a regular convex polychoron, or polytope existing in four dimensions. It is also known as the hexadecachoron. It is one of the six regular convex polychora first described by the Swiss mathematician Ludwig Schläfli in the mid-19th century. Conway calls it an orthoplex for ‘orthant complex’, as well as the entire class of cross-polytopes.

Simple English, eh? That would really demoralize me if I were a non-native speaker.

The 4d cross-polytope

But let’s sidestep the fancy words and think about what the 4d cross-polytope looks like. To draw a cross-polytope in n dimensions, we can draw the n coordinate axes and draw a dot one inch from the origin along each axis in each direction. Then connect each dot to every other one except the opposite one on the same axis. Then erase the coordinate axes.

In 3 dimensions you get this:

It may not look like much, but it’s a perspective picture of the vertices and edges of an octahedron, or 3d cross-polytope.

Puzzle 2. How many line segments going between red dots are in this picture? These are the edges of the 3d cross-polytope.

Puzzle 3. How many triangles with red corners can you see in this picture? These are the triangular faces of the 3d cross-polytope.

Now let’s do the same sort of thing in 4 dimensions! For this we can start with 4 axes in the plane, each at a 45° angle from the next. We can then draw a dot one inch from the origin along each axis in each direction… and connect each dot to each other except the opposite one on the same axis. We get this:

If we then erase the axes, we get this:

This a perspective picture of a 4d cross-polytope!

Puzzle 4. How many line segments going between red dots are in this picture? These are the edges of the 4d cross-polytope.

Puzzle 5. How many triangles with red corners can you see in this picture? These are the triangular 2-dimensional faces of the 4d cross-polytope.

Let’s say that 4d polytope has:

• 0-dimensional vertices,

• 1-dimensional edges,

• 2-dimensional faces, and

• 3-dimensional facets.

In general, the facets of an n-dimensional thing are its (n-1)-dimensional parts, while the parts of every dimension below n are often called faces. But in 4d we have enough words to be completely unambiguous, so let’s use the words as above. And in 3d, let’s use face in its traditional sense, to mean a 2d face.

So, as long as I talk only about 3d and 4d geometry, you can be sure that when I say face I mean a 2-dimensional face. When I say facet, I’ll mean a 3-dimensional face.

Puzzle 6. What shape are the facets of the 4d cross-polytope?

4-cube versus 4d cross-polytope

On top you see the 4-cube. At right, the 4d cross-polytope. Both are projected down to the plane in the same way.

So, the 4d cross-polytope has

2 × 4 = 8

vertices: one centered at each cubical facet of the 4-cube. To see how this works, mentally move the cross-polytope up and put it on top of the 4-cube.

On the other hand, the 4d cross-polytope has

24 = 16

facets: one for each corner of the 4-cube.

And this is a general pattern. As I already observed, the n-dimensional cross-polytope has one vertex in the middle of each facet of the n-cube, and vice versa. For this reason we say they are Poincaré dual to each other, or simply dual. The n-cube has

2 × n

vertices and

2n

facets, but for the n-dimensional cross-polytope it’s the other way around.

Figure credits and more

The picture of the octahedron in cube and cube in octahedron are from Frederick J. Goodman, who has written a book about this stuff called Algebra: Abstract and Concrete.

The other images are on Wikimedia Commons, and all have been released into the public domain except this one:

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I’ll change my article. Since I’ll mainly be talking about 3d and 4d polytopes, I’ll use face to mean a 2d facet in both cases (since in 3d geometry that’s traditional), and facet to mean a 3d facet of a 4d polytope.

P5: 32. This one’s a bit tricky, in that counting the triangles is not completely obvious. I broke it down by taking triangles from each vertex formed by edges 1 through 5 lines apart. But there are no triangles formed by lines 3 apart (since that would connect diagonally opposite vertices). So you end up with 5 (1 apart) + 4 (2 apart) + 0 (3 apart) + 2 (4 apart) + 1 (5 apart) = 12. Then, as in P3, 8 vertices, each with 12 triangles, each shared by 3 vertices = 8 * 12 / 3 = 32.

Namely, for convex polytopes, it’s 0 for even dimensions, but 2 for odd dimensions. But if you include the “greatest” rank (n, the whole object, of which there is always 1) and the “least” rank (-1, the “null face”, of which there is also always 1), then the “abstract Euler characteristic” (to coin a term) is 0 for convex polytopes in all dimensions.

Since a convex polytope is contractible, its Euler characteristic as a topological space is the same as that of a point—namely, 1. (The Euler characteristic of a finite set generalizes the cardinality of a finite set in this way.) To get this answer, you have to count the n-dimensional ‘body’ of the polytope, and it gives If we leave it out, we get the Euler characteristic of the boundary of the polytope, namely

There’s also a modified version of Euler characteristic called ‘reduced Euler characteristic’, where we subtract 1. If we calculate this for a point we get 0, and similarly that’s what we get for a convex polytope.

In future parts of “Symmetry and the Fourth Dimension” will you be discussing the quaternion generators for the 4-D regular polytopes, comparing the SO(4) generators to the SU(2) x SU(2) generators, for example? For the 3-D regular polyhedrons, Coxeter gives three generators, i, j, and a quaternion in the imaginary prime in terms of cosines and sines of p, q, and h. Is there a use for these generators for the 4-D case?

Yes, I’ll be talking about quaternionic descriptions of 4d polytopes. For now, you can read my article Platonic solids in all dimensions. Make sure you make it down to the section ‘Platonic solids and the quaternions’.

I feel a bit guilty about this. I’d planned on a lot more, but then I lost energy. I can’t promise to revive this series. You can however, look at the series of shorter articles that it was supposed to expand on!

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